Calculus Examples

$\int_{0}^{1} \frac{1}{5 + t} d t d x$

Step 1

Let $u = 5 + t$ . Then $d u = d t$ . Rewrite using $u$ and $d$ $u$ .

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Step 1.1

Let $u = 5 + t$ . Find $\frac{d u}{d t}$ .

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Step 1.1.1

Differentiate $5 + t$ .

$\frac{d}{d t} [5 + t]$

Step 1.1.2

By the Sum Rule, the derivative of $5 + t$ with respect to $t$ is $\frac{d}{d t} [5] + \frac{d}{d t} [t]$ .

$\frac{d}{d t} [5] + \frac{d}{d t} [t]$

Step 1.1.3

Since $5$ is constant with respect to $t$ , the derivative of $5$ with respect to $t$ is $0$ .

$0 + \frac{d}{d t} [t]$

Step 1.1.4

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 1$ .

$0 + 1$

Step 1.1.5

Add $0$ and $1$ .

$1$

Step 1.2

Substitute the lower limit in for $t$ in $u = 5 + t$ .

$u_{lower} = 5 + 0$

Step 1.3

Add $5$ and $0$ .

$u_{lower} = 5$

Step 1.4

Substitute the upper limit in for $t$ in $u = 5 + t$ .

$u_{upper} = 5 + 1$

Step 1.5

Add $5$ and $1$ .

$u_{upper} = 6$

Step 1.6

The values found for $u_{lower}$ and $u_{upper}$ will be used to evaluate the definite integral.

$u_{lower} = 5$

$u_{upper} = 6$

Step 1.7

Rewrite the problem using $u$ , $d u$ , and the new limits of integration.

$\int_{5}^{6} \frac{1}{u} d u d x$

Step 2

The integral of $\frac{1}{u}$ with respect to $u$ is $\ln (| u |)$ .

$\ln (| u |)]_{5}^{6} d x$

Step 3

Evaluate $\ln (| u |)$ at $6$ and at $5$ .

$(\ln (| 6 |) - \ln (| 5 |)) d x$

Step 4

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\ln (\frac{| 6 |}{| 5 |}) d x$

Step 5

Simplify.

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Step 5.1

The absolute value is the distance between a number and zero. The distance between $0$ and $6$ is $6$ .

$\ln (\frac{6}{| 5 |}) d x$

Step 5.2

The absolute value is the distance between a number and zero. The distance between $0$ and $5$ is $5$ .

$\ln (\frac{6}{5}) d x$

Step 6